Cyclic Averages of Regular Polygons and Platonic Solids


 Mamuka Meskhishvili
 7 months ago
 Views:
Transcription
1 Cyclic Averages of Regular Polygos ad Platoic Solids Mamuka Meskhishvili Departmet of Mathematics, GeorgiaAmerica High School, 8 Chkodideli Str., Tbilisi 080, Georgia Preprit submitted to RGN Publicatios o 03/06/00 Abstract For a regular polygo P ad Platoic solid T are itroduced the cocept of the cyclic averages. It is show the cyclic averages of equal powers are the same for various P T, but their umber is characteristic of P T. Give the defiitio of a circle sphere by the vertices of P T ad o the base of the cyclic averages are established the commo metrical relatios of P T. MSC. 5M04, 4G05 Keywords ad phrases. Regular polygo, Platoic solid, circle, sphere, locus, sum of like powers, ratioal distaces problem Article type: Research article Itroductio Cosider a fiite set of poits i the plae space, the locus of poits such that the sum of the squares of distaces to the give poits is costat, is a circle sphere, whose ceter is at the cetroid of the give poits [, ]. Deote by Md, d,..., d, L a arbitrary poit i the plae space of a regular polygo Platoic solid of distaces d, d,..., d to the vertices A, A,..., A, the: d i = R + L, where R is the radius of circumscribed circle sphere of the regular polygo Platoic solid ad L is the distace betwee the poit M ad the cetroid O. For a equilateral triagle ad a arbitrary poit Md, d, d 3, L i the plae of the triagle the symmetric equatio exists 3d 4 + d 4 + d a 4 = d + d + d 3 + a,
2 where a is the side of the triagle [3, 4]. The arbitrary poit always is cosidered i the plae of the regular polygo, ad i the space of the Platoic solid, respectively. From relatios ad follows: 3 d i = 3R + L, 3 d 4 i = 3 R + L + R L. For a give equilateral triagle, the the side a as well as the circumradius R are fixed so that, Theorem.. The locus of poits such that is a circle, ceter of which is the cetroid. 3 d 4 i = cost As we see, the distaces are cosidered to the secod ad the fourth powers. Naturally, we are iterested to kow what happes if we cosider the distaces of higher powers. Prelimiaries For a equilateral triagle the expressio 3 d 6 i, cotais α the agle betwee R ad L, so the locus is ot a circle, but for a square case the aswer is surprisig: the locus of poits such that is a circle. d 6 i = cost Geerally, the locus is a circle sphere if ad oly if the sum of power distaces ca be expressed i terms of L ad some fixed elemet with legth of a give regular polygo Platoic solid. The fixed elemet is possible to express i terms of R, so we deote such sums by the symbol [R,L], or m [R,L] to idicate the like powers of the distaces. Deote by P R ad T R a regular polygo ad Platoic solid, respectively, with a umber of the vertices ad with circumscribed radius R. The value of the [R,L] remais costat whe the poit M moves o the circle CO, L sphere SO, L. So, Defiitio.. [R,L] is the sum of like powers of the distaces d,..., d from a arbitrary poit Md,..., d, L to the vertices P R T R the value of which is costat for ay poit of the CO, L SO, L.
3 It is clear, the sum of odd power cotais radicals ad ever will be [R,L]. For establishig commo properties of the P ad T discussig average of [R,L] prefered S m = m. [R,L] is much Defiitio.. The cyclic averages S m average of the sum m [R,L]. S m [] of a regular polygo Platoic solid is the We call such averages as the cyclic averages, because as we prove for fixed R ad L the cyclic averages of equal powers of differet P T are the same if they exist: if. S m = S m, O the other had for ay give P the umber of S m as well as m [R,L] is defied uiquely, so the umber of the cyclic averages is characteristic of the regular polygo. For example for a regular 3go exists cyclic averages: while for a regular 4go 3 cyclic averages: They are i relatios: S 3 ad S 4 3, S 4, S4 4 ad S 6 4. S 3 = S 4 ad S 4 3 = S 4 4. To demostrate the efficiecy of cyclic averages the aalogue of the relatio will be obtaied for the square. Firstly we tur i terms of R ad the cyclic averages S 3, S4 3 : the replace with d 4 + d4 + d R 4 = d + d + d R, S 3 = S 4, S4 3 = S 4 4 ad R = a ; we get Theorem.. For a arbitrary poit Md,..., d 4, L i the plae of a square: 4d 4 + d 4 + d d a 4 = d + d + d 3 + d 4 + a, where a is the side of the square. 3
4 3 Circle as Locus of Costat [R,L] Sums Theorem 3.. For a arbitrary poit Md, d,..., d, L i the plae of regular polygo P R: i= d m i where m =,...,. [ = R + L m + First we eed to prove two lemmas. m k= m k k k R k L k R + L m k ], Lemma 3.. For arbitrary positive itegers m ad, such that m <, the followig coditio k= is satisfied, where α is a arbitrary agle. Deote The real part of T is cos m α k π = 0 T = e imα + e imα π + e imα π + + e imα π. ReT = k= cos m α k π. The formula of the sum of geometric progressio gives T = e imα + e im π + e im π + + e im π e im π = cos πm + i πm =. π imα e im = e, e im π Sice m <, e im π. So T = 0, i.e. ReT = 0, which proves Lemma 3.. Remark 3.. If m, the sum always cotais α. Lemma 3.. For arbitrary positive itegers m ad, such that m < ad for a arbitrary agle α the followig coditios are satisfied: if m is odd cos m α k π = 0; k= 4
5 if m is eve cos m α k π = k= mm m. Whe m is odd, usig the powerreductio formula for cosie cos m θ = m m k=0 m cos m kθ, k we obtai cos m α k π k= = cos m α + cos m α π + + cos m α π [ m = m m m cos mα + cosm α + + m 0 m + cos m α π m + cosm α π 0 m + m cos α π + + = m m cos m α π 0 [ m m + + m cos mα + cos m m + cosm cos α π cos α + α π ] + + cos m α π α π 0 m + cosm α+cosm α π + + cosm α π + m + m cos α + cos α π + + cos α π ]. Sice m <, from Lemma 3. it follows that each sum equals zero, which proves the first part of Lemma 3.. Whe m is eve, the powerreductio formula for cosie is cos m θ = m m m + m m k=0 5 m cos m kθ. k
6 Aalogously to the case with odd m, the sum of the secod addeda vaishes, ad sice the umber of the first addeda is, the total sum equals m m m, which proves Lemma 3.. Proof of Theorem 3.. We itroduce the ew otatios A = R + L ad B = RL. The i= π m d m i = A B cos α m + A B cos α If m =, by Lemma 3. we have i= Therefore d m i = A B cos α If m >, we have i= d m i + A B cos π m α + + A B cos π m α. π + A B cos α + + A B cos π α = A. m = A m A m B m + A m B m A m 3 B 3 3 m ± B m m S = R + L. π cos α + cos α + + cos π α cos α + cos π α + + cos π α cos 3 α + cos 3 π α + + cos 3 π α + cos m α + cos m π α + + cos m π α. Accordig to Lemma 3., all sums with the egative sig vaish ad oly the sums with the positive sig remai. 6
7 i= If m is eve d m i = A m + = If m is odd i= d m i = A m + = m A m B cos α + cos π α + + cos π α + m + B m cos m α + cos m π m α m m A m + A m k B k k k k. k + A m + k= m m m A m B cos α + cos π m k= AB m m A m k B k k k + + cos m π α α + + cos π α + cos m α + cos m π α + + cos m π α k. k Usig the floor fuctio the iteger part, the obtaied results ca be combied ito a sigle formula as follows i= which proves the theorem. d m i From Theorem 3. each sums = A m + i= m k= m A m k B k k k d m i, where m =,,..., k, k are the [R,L] sums. Begiig from the m all sums of power distaces cotai α Remark 3.. For example for P 3 the sums cotai:  cos 3α, if m = 3, 4, 5;  cos 3α ad cos 6α, if m = 6, 7, 8;  cos 3α, cos 6α ad cos 9α, if m = 9, 0,. 7
8 Geerally for m the sums i= d m i such sums is beyod the scope of this article. cotai cosie of the multiples of α. The study of Therefore for P exist a umber of [R,L] sums ad if they are costat the locus for each case is a circle: Theorem 3.. The locus of poits such that the sum of the mth power of the distaces to the vertices of a give P R is costat is a circle, if i= whose ceter is the cetroid of the P R. Remark If  If i= i= d m i d m i d m i > R m, where m =,,...,, = R m, the locus is the cetroid of the polygo. < R m, the locus is the empty set. 4 Cyclic Averages of Regular Polygos The properties of the cyclic average are as follows: Property 4.. Each regular go has a umber of cyclic averages S, S 4,..., S. Property 4.. For fixed R ad L, the cyclic averages of equal powers of differet regular gos are the same: S 3 = S 4 = S 5 = S 6 =, S 4 3 = S 4 4 = S 4 5 = S 4 6 =, S 6 4 = S 6 5 = S 6 6 =, S 8 5 = S 8 6 =. Property 4.3. Ay relatios i terms of the cyclic averages S m, the circumscibed radius R ad the distace L, which are satisfied for a regular go, are at the same time satisfied for ay regular go, where, i.e. S m ca be replaced by S m. Elimiate L from the relatios of Theorem 3. we obtai: 8
9 Theorem 4.. For ay regular go: where m =,...,. S m = S m + I terms of S ad S 4 : m k= Theorem 4.. For ay regular go: where m = 3,...,. m S m = S m + k= k m k m k k k k k The first two relatios of Theorem 3. implies: Theorem 4.3. For ay regular go: R = so L = S ± S The poits o the circumscribed circle satisfy R k S R k S m k, S 4 S k S m k, 3S S 4 3S S 4 3S = S 4, Theorem 4.4. For ay poit o the circumscribed circle of the regular go: 4. Equilateral triagle There are cyclic averages: 3 i= d i = d 4 i. i= S 3 = 3 d + d + d 3 = R + L, I geeral case from Theorem 4., for 3 [5],. S 4 3 = 3 d4 + d 4 + d 4 3 = R + L + R L. S 4 + 3R 4 = S + R. Deote by the symbol a,b,c the area of a triagle whose sides have legths a, b, c. The solutio of the system of the cyclic averages is: 9
10 Theorem 4.5. For ay poit Md, d, d 3, L ad P 3 R d = d, d = 3R + L d ± 4 3 R,L,d, d 3 = 3R + L d 4 3 R,L,d. ad For P 3 3S S 4 = d + d + d 3 3 d 4 + d 4 + d 4 3 = 6 3 d,d,d 3, R = d + d + d 3 ± d,d,d 3, L = d + d + d d,d,d 3. For ay poit o the circumscribed circle, follows the area d,d,d 3 should be zero. Ideed for the largest distace d 3 = d + d holds. 4. Square There are 3 cyclic averages: S 4 = 4 d + d + d 3 + d 4 = R + L, S 4 4 = 4 d4 + d 4 + d d 4 4 = R + L + R L, S 6 4 = 4 d6 + d 6 + d d 6 4 = R + L 3 + 6R L R + L. From Theorems 4. ad 4. Theorem 4.6. For ay regular go, where 4: S 6 = S S + 3R 5R 4, S 6 = S 3S 4 S. From Theorem 4.6 follows: 8d 6 + d 6 + d d d + d + d 3 + d 4 3 = 6d + d + d 3 + d 4d 4 + d 4 + d d 4 4, which is equivalet to 3d + d d 3 d 4d + d 3 d d 4d + d 4 d d 3 = 0. 0
11 So holds. d + d 3 = d + d 4 Obtaied relatio has geeralizatio for regular go. If is eve for the diametrically opposed vertices: Theorem 4.7. For ay regular go, with eve umber of vertices = k: d + d +k = d + d +k = = d k + d k = R + L. Theorem 4.7 simplifies the system of the cyclic averages: S R = S 4 + R, d + d 3 = d + d 4; which is aalogue to systems obtaied i [6, 7]. Moreover i terms of R ad L, we get: The solutio of which is: d + d 3 = d + d 4 = R + L, d d 3 + d d 4 = R 4 + L 4. Theorem 4.8. For ay poit Md, d, d 3, d 4, L ad P 4 R: d = d, d = R + L ± 4 R,L,d, d 3 = R + L d, d 4 = R + L 4 R,L,d. For P 4 3S S = [ ] 3d + d + d 3 + d 6 4 8d 4 + d 4 + d d 4 4 = 4 d, d,d 3 = 4 d, d 3,d 4, ad R = 4 d + d 3 ± d, d,d 3 = 4 d + d 4 ± d, d 3,d 4, L = 4 d + d 3 d, d,d 3 = 4 d + d 4 d, d 3,d 4. For ay poit o the circumscribed circle the areas d, d,d 3 ad d, d 3,d 4 should be zero. Ideed, if the poit o the mior arc A A are satisfied d + d = d 3 ad d + d 4 = d 3.
12 4.3 Regular Petago, Hexago ad Heptago There are 4, 5 ad 6 cyclic averages for the P 5, P 6 ad P 7 cases, respectively: S 5 = S 6 = S 7 = R + L, S 4 5 = S 4 6 = S 4 7 = R + L + R L, S 6 5 = S 6 6 = S 6 7 = R + L 3 + 6R L R + L, S 8 5 = S 8 6 = S 8 7 = R + L 4 + R L R + L + 6R 4 L 4, S 0 6 = S 0 7 = R + L 5 + 0R L R + L R 4 L 4 R + L, S 7 = R + L R L R + L R 4 L 4 R + L + 0R 6 L 6. These systems are simplified for the regular hexago case oly. The vertices A, A 3, A 5 ad A, A 4, A 6 form two equilateral triagles, so they satisfy two cyclic relatios for P 3. Geerally for go if divisible by 3: Theorem 4.9. For ay regular go, if = 3l d + d +l + d +l = = d l + d l + d 3l = 3R + L, d 4 + d 4 +l + d4 +l = = d4 l + d4 l + d4 3l = 3 R + L + R L. The Theorem 4.7 ad Theorem 4.9 simplify the system of the cyclic averages for the regular hexago: d + d 4 = d + d 5 = d 3 + d 6 = R + L, d + d 3 + d 5 = d + d 4 + d 6 = 3R + L, d 4 + d d 4 5 = d 4 + d d 4 6 = 3 R + L + R L. By usig these relatios, we get explicit expressios for distaces: Theorem 4.0. For ay poit Md, d,..., d 6, L ad P 6 R: d = d, d = R + L + d ± 4 3 R,L,d, d 3 = 3R + 3L d ± 4 3 R,L,d d 4 = R + L d, d 5 = 3R + 3L d 4 3 R,L,d d 6 = R + L + d 4 3 R,L,d.,,
13 For P 6 : ad d 3S S 4 = 3 + d + + d 6 d 4 + d4 + + d d + d 3 + d 5 d 4 + d d 4 5 = 3 = 6 3 d,d 3,d 5 = 6 3 d,d 4,d 6 R = d + d 3 + d 5 ± d,d 3,d 5 = d + d 4 + d 6 ± d,d 4,d 6, L = d + d 3 + d d,d 3,d 5 = d + d 4 + d d,d 4,d 6. For ay poit o the circumscribed circle the area d,d 3,d 5 as well as d,d 4,d 6 vaishes. Ideed if the poit o the mior are A A : d + d 3 = d 5 ad d + d 6 = d Regular Octago, Noago ad Decago There are 8, 9 ad 0 cyclic averages for the P 8, P 9 ad P 0 cases, respectively. The cyclic averages from the secod to the twelfth powers are the same as for regular heptago, so we write oly ew oes: S 4 8 = S 4 9 = S 4 0 = R + L 7 + 4R L R + L 5 + 0R 4 L 4 R + L R 6 L 6 R + L, S 6 9 = S 6 0 = R + L R L R + L R 4 L 4 R + L R 6 L 6 R + L + 70R 8 L 6, S 8 0 = R + L 9 + 7R L R + L R 4 L 4 R + L 5 All three cases = 8, 9, 0 admit further simplificatios. For P 8 Theorem 4.7 gives: + 680R 6 L 6 R + L R 8 L 8 R + L, d + d 5 = d + d 6 = d 3 + d 7 = d 4 + d 8 = R + L. The vertices A, A 3, A 5, A 7 ad A, A 4, A 6, A 8 form two squares, so they satisfy additioal cyclic relatios for P 4. Geerally, if is divisible by 4: Theorem 4.. For ay regular go, if = 4p: d 4 + d 4 +p + d 4 +p + d 4 +3p = = d 4 p + d 4 p + d 4 3p + d 4 4p = 4 R + L + R L, d 6 + d 6 +p + d 6 +p + d 6 +3p = = d 6 p + d 6 p + d 6 3p + d 6 4p = 4 R + L 3 + 6R L R + L. 3
14 For P 9 Theorem 4.9 gives: d + d 4 + d 7 = d + d 5 + d 8 = d 3 + d 6 + d 9 = 3R + L, d 4 + d d 4 7 = d 4 + d d 4 8 = d d d 4 9 = 3 R + L + R L. For P 0, from Theorem 4.7: d + d 6 = d + d 7 = d 3 + d 8 = d 4 + d 9 = d 5 + d 0 = R + L. The vertices A, A 3, A 5, A 7, A 9 ad A, A 4, A 6, A 8, A 0 form two regular petagos, so they satisfy additioal cyclic relatios for P 5. Geerally, if is divisible by 5: Theorem 4.. For ay regular go, if = 5t d + d +t + d +t + d +3t + d +4t = = d t + d t + d 3t + d 4t + d 5t = 5R + L, d 4 + d 4 +t + d 4 +t + d 4 +3t + d 4 +4t = = d 4 t + d 4 t + d 4 3t + d 4 4t + d 4 5t = 5 R + L + R L, d 6 + d 6 +t + d 6 +t + d 6 +3t + d 6 +4t = = d 6 t + d 6 t + d 6 3t + d 6 4t + d 6 5t = 5 R + L 3 + 6R L R + L, d 8 + d 8 +t + d 8 +t + d 8 +3t + d 8 +4t = = d 8 t + d 8 t + d 8 3t + d 8 4t + d 8 5t = 5 R + L 4 + R L R + L + 6R 4 L 4. Summarize the obtaied results, we coclude: every regular go has a umber of cyclic averages, but if is the composite umber we have additioal relatios for the distaces, which are obtaied from the cyclic averages of the go, where is divisible of. 5 Ratioal Distaces Problem. Solutio for = 4 Is there a poit all of whose distaces to the vertices of the uit polygo are ratioal? The problem has a log history especially for the case of square. A extesive historical review is give i [6 8]. For case of a equilateral triagle aswer is positive [9]. Accordig [0] ope problems are i followig cases = 4, 6, 8, ad 4. For = 6 oly trivial poit is kow the cetroid of the uit hexago. By Theorem 4.3 the side a of the regular go is: a si π For the uit icositetrago = 4: si π 4 = = S ± S 3S S 4. 4 ± 3S 4 S 4 4 S 4 4. S 4 4
15 The right side is the root of the fourth degree polyomial equatio with ratioal coefficiets: 8 S 4 S x 4 4S x + = 0, thus it is the algebraic umber of degree 4. O the other had si π 4 = + 3, is the algebraic umber of degree > 4 []. So, Theorem 5.. There is ot a poit i the plae that is at ratioal distaces from the vertices of the uit regular 4go. For positive aswers for the P 4 ad P 6 cases the ecessary coditios are the ratioalities of the equal areas:  d, d,d 3 = d, d 3,d 4, if = 4;  3 d,d 3,d 5 = 3 d,d 4,d 6, if = 6. 6 Sphere as Locus of Costat [R,L] Sums For regular polygos with differet vertices the umber of the [R,L] sums are differet too. As we see, ulike the plae case, dual Platoic solids have the same umber of the [R,L] sums: regular tetrahedro octahedro ad cube [R,L], 4 [R,L], icosahedro ad dodecahedro [R,L], 4 [R,L], 6 [R,L], To prove these, we cosider each platoic solid separately. cetered at the origi ad use simple Cartesia coordiates. 6. Regular Tetrahedro The coordiates of the vertices T 4 R: [R,L], 4 [R,L], 6 [R,L], 8 [R,L], 0 [R,L]. A, c, ±c, ±c, A 3,4 c, ±c, c ad R = 3 c. I all cases, we cosider solids Cosider a arbitrary poit i space Md, d, d 3, d 4, L with the coordiates: x, y, z. The distace betwee M ad the cetroid O of the tetrahedro: L = x + y + z. 5
16 The, d, = x c + y c + z c = R + L + c x y z, d 3,4 = x + c + y c + z ± c = R + L + cx y ± z, d 4 i = R + L + c x y z + R + L + cx y ± z = 4R + L + 4c x + y + z + x + y + z + x y + z + x + y z = 4 R + L R L. If for T 4 R: the d 4 i > 4R 4, Theorem 6.. The locus of poits i the space such that the sum of the fourth power of the distaces to the vertices of a give regular tetrahedro is costat is a sphere whose ceter is the cetroid of the tetrahedro. Remark If  If d 4 i = 4R4 the locus is the cetroid. d 4 i < 4R4 the locus is the empty set. The sums of the distaces of the power more tha 4 cotai x, y ad z like α for the plae case, so for T 4 oly the sums of the secod ad fourth powers are [R,L] sums. 6. Octahedro ad Cube The coordiates of the vertices of the octahedro T 6 R: A, ±c, 0, 0, A 3,4 0, ±c, 0, A 5,6 0, 0, ±c ad R = c. For a arbitrary poit P d, d,..., d 6, L: d, = R + L ± Rx, d 3,4 = R + L ± Ry, d 5,6 = R + L ± Rz. 6
17 Begiig from T 6 each Platoic solid has except tetrahedro diametrically opposed vertices, so for them is satisfied Theorem 4.7. For the sums of the fourth ad sixth powers: For the cube T 8 R: ad R = 3 c. 6 d 4 i = R + L ± Rx + R + L ± Ry ± R + L ± Rz = 4 R + L R L, 6 d 6 i = 6R + L 3 + 4R + L R x + y + z = 6 R + L 3 + 4R L R + L. The distaces from the P d, d,..., d 8, L: A, c, c, c, A 3,4 ±c, ±c, c, A 5,6 ±c, c, ±c, A 7,8 c, ±c, ±c d, = R + L ± cx + y + z, d 5,6 = R + L cx + z y, d 3,4 = R + L cx + y z, d 7,8 = R + L ± cx y z. The vertices A, A 3, A 5, A 7 ad A, A 4, A 6, A 8 form two regular tetrahedros, so they satisfy the regular tetrahedro relatios. Theorem 6.. For a arbitrary poit i the space, the sum of the quadruple of the distaces to the vertices of the cube which lie o parallel faces ad are edpoits of skew face diagoals, satisfies d k = d 4 k = d k = 4R + L, d 4 k R = 4 + L R L. Remark 6.. These quadruples do ot cotai the distaces to diametrically opposed vertices. Thus, 8 d 4 i = 8 R + L R L. 8 d 6 i = R + L ± cx + y + z 3 + R + L cx + y z 3 7
18 the + R + L cx + z y 3 + R + L cy + z x 3 = 8R + L 3 + 4R + L c x + y + z + x + y z + x y + z + x y z = 8 R + L 3 + 4R L R + L. If for T 6 R ad T 8 R is satisfied i= d m i > R m, = 6, 8; Theorem 6.3. The locus of poits i the space such that the sum of the sixth fourth power of distaces to the vertices of a give octahedro cube is costat is a sphere whose ceter is the cetroid of the octahedro cube. Remark If  If d m i d m i = R m the locus is the cetroid. < R m the locus is the empty set. 6.3 Icosahedro ad Dodecahedro The coordiates of the vertices of icosahedro T R: A, 0, ±c, ±cϕ, A 3,4 0, c, ±cϕ, A 5,6 ±c, ±cϕ, 0, A 7,8 ±c, cϕ, 0, A 9,0 ±cϕ, 0, ±c, A, ±cϕ, 0, c, where ϕ is the golde ratio ϕ = + 5 ad R = c + ϕ. For a arbitrary poit P d, d,..., d, L: d, = R + L cy + zϕ, d 5,6 = R + L cx + yϕ, d 9,0 = R + L cz + xϕ, d 3,4 = R + L ± cy zϕ, d 7,8 = R + L cx yϕ, d, = R + L ± cz xϕ. 8
19 The d 4 i = d 6 i = d 4 i + 8 d 4 i d 4 i = 4R + L + 4c y + z ϕ + 4R + L + 4c x + y ϕ + 4R + L + 4c z + x ϕ = R + L + 6c + ϕ x + y + z = R + L R L. d 6 i + For the sum of the eight power 8 d 6 i = 4R + L c R + L y + z ϕ d 6 i + 4R + L c R + L x + y ϕ + 4R + L c R + L z + x ϕ = R + L 3 + 4R L R + L. d 8 i = 4R + L c R + L y + z ϕ + 64c 4 y 4 + z 4 ϕ 4 + 6y z ϕ, Because d 8 i = R + L c R + L x + y + z + ϕ + 64c 4 x 4 + y 4 + z 4 + ϕ 4 + 6x y + x z + y z ϕ, + ϕ 4 = 3ϕ ad ϕ = 5 + ϕ, d 8 i = R + L 4 + 8R L R + L R4 L 4. d 0 i = 4R + L R + L 3 c y + zϕ + y zϕ + 60R + L c 4 y + zϕ 4 + y zϕ 4 = 4R + L R + L 3 c y + zϕ + 30R + L c 4 y 4 + z 4 ϕ 4 + 6y z ϕ. d 0 i = R + L R + L 3 c x + y + z + ϕ 9
20 + 30R + L c 4 + ϕ 4 x 4 + y 4 + z 4 + 3ϕ x y + x z + y z = R + L R L R + L 3 + 6R 4 L 4 R + L. The vertices of the dodecahedro T 0 R divide ito two groups, the vertices A, A,..., A 8 which form a cube ad other vertices A 9, A 0,..., A 0. The the coordiates: ad R = 3 c. A, c, c, c, A 3,4 ±c, ±c, c, A 5,6 ±c, c, ±c, A 7,8 c, ±c, ±c, A 9,0 0, ± c ϕ, ±cϕ, A, 0, c, ϕ, ±cϕ A 3,4 ± c ϕ, ±cϕ, 0, A 5,6 c, ϕ, ±cϕ, 0 A 7,8 ± cϕ, 0, ± c, A 9,0 ± cϕ, 0, c. ϕ ϕ Cosider a arbitrary poit P d, d,..., d 0, L. For the distaces d, d,..., d 8 we use the respective distaces of the cube, ad for others: y y d 9,0 = R + L c ϕ + zϕ, d, = R + L ± c ϕ zϕ, x x d 3,4 = R + L c ϕ + yϕ, d 5,6 = R + L ± c ϕ yϕ, z z d 7,8 = R + L c ϕ + xϕ, d 9,0 = R + L ± c ϕ xϕ, 0 0 d 4 i = 8R + L R L + 0 d 4 i 9 = 8R + L R L + R + L + 6c x + y + z ϕ + ϕ = 0 R + L R L. d 6 i = 8R + L 3 + 3R L R + L + R + L 3 + 3R + L 6c x + y + z ϕ + ϕ = 0 R + L 3 + 4R L R + L. 8 d 8 i = R + L ± cx + y + z 4 + R + L cx + y z 4 + R + L cx + z y 4 + R + L ± cx y z 4 = 8R + L R L R + L R4 L 4 + 8x y + 8x z + 8y z, 0
21 the 0 d 8 i = R + L R + L L 3c c 4 x 4 + y 4 + z 4 ϕ 4 + ϕ 4 + 6x y + 6x z + 6y z = R + L R + L R L R4 7x 4 + y 4 + z 4 + 6x y + 6x z + 6y z, 0 d 8 i = 0 R + L 4 + 8R L R + L R4 L 4. Like T, maximal power for T 0 which depeds o R ad L oly is 0. Ideed, d 0 i = R + L ± cx + y + z 5 + R + L cx + y z 5 + R + L cx + z y 5 + R + L ± cx y z 5 = 8R + L R + L 3 c L + 30R L c 4 x 4 + y 4 + z 4 + x y + x z + y z, d 0 i = R + L R + L 3 c L ϕ + ϕ + 30R + L c 4 ϕ 4 + ϕ4 x 4 + y 4 + z 4 + 6x y + x z + y z, d 0 i = 0 R + L R L R + L 3 + 6R 4 L 4 R + L. If for T R ad T 0 R is satisfied i= d m i > R m, =, 0, Theorem 6.4. The locus of poits i the space such that the sum of the mth power of distaces to the vertices of a give icosahedro dodecahedro is costat is a sphere, whe m =,, 3, 4 ad 5. The ceter of the sphere is the cetroid of the icosahedro dodecahedro.
22 Remark If = R m the locus is the cetroid.  If i= i= d m i d m i < R m the locus is the empty set. 7 Cyclic Averages of Platoic Solids Summurize the obtaied results, i terms of the cyclic averages: Theorem 7.. The cyclic averages of the Platoic solids are the followig: S [4] = S [6] = S [8] = S [] = S [0] = R + L, S 4 [4] = S 4 [6] = S 4 [8] = S 4 [] = S4 [0] = R + L R L, S 6 [6] = S 6 [8] = S 6 [] = S6 [0] = R + L 3 + 4R L R + L, S 8 [] = S8 [0] = R + L 4 + 8R L R + L R4 L 4, S 0 [] = S 0 [0] = R + L R L R + L 3 + 6R 4 L 4 R + L. Elimiate L ad R from the relatios, we obtai direct relatios amog the cyclic averages of the Platoic solids. Theorem 7.. For each Platoic solid = 4, 6, 8,, 0: S 4 [] R4 = S [] + 3 R. This result for regular simplicial ad regular polytopic distaces is obtaied i [] ad [3], respectively. Theorem 7.3. For each Platoic solid, except the tetrahedro = 6, 8,, 0: S 6 [] = S [] S [] + R 8R 4, S 6 [] = S 4 [] 3S [] S []. Theorem 7.4. For the icosahedro ad the dodecahedro =, 0: S 8 [] S [] 4 = 8R S [] R S [] + 5 R S [] R, S 0 [] S [] 5 = 8R S 5 [] S [] R 3 S [] + R S [] R S 8 [] = 5 9S 4 S 0 [] = S [] S4 [] [] + S 4 [] S 9S [] 8S []. [] 6S [] 4,,
23 Like the plae cases, i some space cases we have additioal relatios. Each Platoic solid, except the tetrahedro satisfies Theorem 4.7 ad for the cube ad the dodecahedro Theorem 6.. For the radius of the circumscribed sphere ad the distace betwee the poit ad the cetroid: Theorem 7.5. For each Platoic solid = 4, 6, 8,, 0: R = S [] ± 4S [] 3S 4 [], L = S [] 4S [] 3S 4 []. so The poits o the circumscribed sphere satisfy 4S [] = 3S 4 [], Theorem 7.6. For ay poit o the circumscribed sphere of each Platoic solid =4, 6, 8,, 0: 4 d i = 3 d 4 i. 8 Coclusio I the preset paper, we itroduce the [R,L] sums ad defie the cyclic averages of the regular polygos ad the Platoic solids. We prove the mai property of the cyclic averages the equality of them for various regular polygos ad Platoic solids. By meas of the cyclic averages the distaces of a arbitrary poit to the vertices of the regular polygos the plae case ad the Platoic solids the space case are ivestigated. It is foud all cases of costat sum of like powers of the distaces, whe the locus is a circle a sphere, established the metrical relatios for regular polygos Platoic solids, which were kow i special cases oly. Ratioal distaces problem solved for the = 4 case. Ackowledgemet The author would like to thak Georgia Youg Scietists Uio GYSU. This research was fuded by GYSU. Cotract o
24 Refereces [] T. M. Apostol ad M. A. Matsakaia, Sums of squares of distaces. Math Horizos 9. 00,. [] T. M. Apostol ad M. A. Matsakaia, Sums of squares of distaces i mspace. America Mathematical Mothly 0 003, o. 6, [3] M. Garder, Mathematical circus. Mathematical Associatio of America, 99. [4] B. J. McCarti, Mysteries of the Equilateral Triagle. Hikari Ltd., Ruse, 00. [5] J. Beti, Regular polygoal distaces. The Mathematical Gazette 8 997, o. 49, [6] R. K. Guy, Usolved Problems i Number Theory. SprigerVerlag, New York Berli, 98. [7] T. G. Berry, Poits at ratioal distace from the corers of a uit square. Aali della Scuola Normale Superiore di Pisa. Classe di Scieze. Serie IV 7 990, o. 4, [8] R. K. Guy, Tilig the square with ratioal triagles. Number Theory ad Applicatios, RA Molli ed., NATO Adv. Study Ist. Ser. C , [9] R. Barbara ad A. Karam, The ratioal distace problem for equilateral triagles. Commuicatios i Mathematics ad Applicatios 9 08, o., [0] R. Barbara, Poits at ratioal distace from the vertices of a uit polygo. Bulleti of the Iraia Mathematical Society , o., [] P. Tagsupphathawat, Algebraic trigoometric values at ratioal multipliers of π. Acta et Commetatioes Uiversitatis Tartuesis de Mathematica 8 04, o., 9 8. [] J. Beti, Regular simplicial distaces. The Mathematical Gazette , p. 06. [3] P.S. Park, Regular polytopic distaces. Forum Geometricorum 6 06,
CALCULUS BASIC SUMMER REVIEW
CALCULUS BASIC SUMMER REVIEW NAME rise y y y Slope of a o vertical lie: m ru Poit Slope Equatio: y y m( ) The slope is m ad a poit o your lie is, ). ( y SlopeItercept Equatio: y m b slope= m yitercept=
More informationMath 155 (Lecture 3)
Math 55 (Lecture 3) September 8, I this lecture, we ll cosider the aswer to oe of the most basic coutig problems i combiatorics Questio How may ways are there to choose a elemet subset of the set {,,,
More information62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 +
62. Power series Defiitio 16. (Power series) Give a sequece {c }, the series c x = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + is called a power series i the variable x. The umbers c are called the coefficiets of
More informationChapter 4. Fourier Series
Chapter 4. Fourier Series At this poit we are ready to ow cosider the caoical equatios. Cosider, for eample the heat equatio u t = u, < (4.) subject to u(, ) = si, u(, t) = u(, t) =. (4.) Here,
More informationA Recurrence Formula for Packing HyperSpheres
A Recurrece Formula for Packig HyperSpheres DokeyFt. Itroductio We cosider packig of D hyperspheres of uit diameter aroud a similar sphere. The kissig spheres ad the kerel sphere form cells of equilateral
More informationCALCULATION OF FIBONACCI VECTORS
CALCULATION OF FIBONACCI VECTORS Stuart D. Aderso Departmet of Physics, Ithaca College 953 Daby Road, Ithaca NY 14850, USA email: saderso@ithaca.edu ad Dai Novak Departmet of Mathematics, Ithaca College
More informationStanford Math Circle January 21, Complex Numbers
Staford Math Circle Jauary, 007 Some History Tatiaa Shubi (shubi@mathsjsuedu) Complex Numbers Let us try to solve the equatio x = 5x + x = is a obvious solutio Also, x 5x = ( x )( x + x + ) = 0 yields
More informationMath 451: Euclidean and NonEuclidean Geometry MWF 3pm, Gasson 204 Homework 3 Solutions
Math 451: Euclidea ad NoEuclidea Geometry MWF 3pm, Gasso 204 Homework 3 Solutios Exercises from 1.4 ad 1.5 of the otes: 4.3, 4.10, 4.12, 4.14, 4.15, 5.3, 5.4, 5.5 Exercise 4.3. Explai why Hp, q) = {x
More informationSEQUENCES AND SERIES
Sequeces ad 6 Sequeces Ad SEQUENCES AND SERIES Successio of umbers of which oe umber is desigated as the first, other as the secod, aother as the third ad so o gives rise to what is called a sequece. Sequeces
More informationOn a Problem Regarding the nsectors of a Triangle
Forum Geometricorum Volume 5 (2005) 47 52. FORUM GEOM ISSN 15341178 O a Problem Regardig the Sectors of a Triagle Bart De Bruy Abstract. Let be a triagle with vertices A, B, C ad agles α = BAC, β = ÂBC,
More informationMAT 271 Project: Partial Fractions for certain rational functions
MAT 7 Project: Partial Fractios for certai ratioal fuctios Prerequisite kowledge: partial fractios from MAT 7, a very good commad of factorig ad complex umbers from Precalculus. To complete this project,
More informationThe Random Walk For Dummies
The Radom Walk For Dummies Richard A Mote Abstract We look at the priciples goverig the oedimesioal discrete radom walk First we review five basic cocepts of probability theory The we cosider the Beroulli
More information(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3
MATH 337 Sequeces Dr. Neal, WKU Let X be a metric space with distace fuctio d. We shall defie the geeral cocept of sequece ad limit i a metric space, the apply the results i particular to some special
More informationThe picture in figure 1.1 helps us to see that the area represents the distance traveled. Figure 1: Area represents distance travelled
1 Lecture : Area Area ad distace traveled Approximatig area by rectagles Summatio The area uder a parabola 1.1 Area ad distace Suppose we have the followig iformatio about the velocity of a particle, how
More informationOn Some Properties of Digital Roots
Advaces i Pure Mathematics, 04, 4, 9530 Published Olie Jue 04 i SciRes. http://www.scirp.org/joural/apm http://dx.doi.org/0.436/apm.04.46039 O Some Properties of Digital Roots Ilha M. Izmirli Departmet
More informationThe ztransform. 7.1 Introduction. 7.2 The ztransform Derivation of the ztransform: x[n] = z n LTI system, h[n] z = re j
The Trasform 7. Itroductio Geeralie the complex siusoidal represetatio offered by DTFT to a represetatio of complex expoetial sigals. Obtai more geeral characteristics for discretetime LTI systems. 7.
More informationComplex Numbers Solutions
Complex Numbers Solutios Joseph Zoller February 7, 06 Solutios. (009 AIME I Problem ) There is a complex umber with imagiary part 64 ad a positive iteger such that Fid. [Solutio: 697] 4i + + 4i. 4i 4i
More informationLesson 10: Limits and Continuity
www.scimsacademy.com Lesso 10: Limits ad Cotiuity SCIMS Academy 1 Limit of a fuctio The cocept of limit of a fuctio is cetral to all other cocepts i calculus (like cotiuity, derivative, defiite itegrals
More informationMAT1026 Calculus II Basic Convergence Tests for Series
MAT026 Calculus II Basic Covergece Tests for Series Egi MERMUT 202.03.08 Dokuz Eylül Uiversity Faculty of Sciece Departmet of Mathematics İzmir/TURKEY Cotets Mootoe Covergece Theorem 2 2 Series of Real
More informationMa 530 Introduction to Power Series
Ma 530 Itroductio to Power Series Please ote that there is material o power series at Visual Calculus. Some of this material was used as part of the presetatio of the topics that follow. What is a Power
More informationChapter 2. Periodic points of toral. automorphisms. 2.1 General introduction
Chapter 2 Periodic poits of toral automorphisms 2.1 Geeral itroductio The automorphisms of the twodimesioal torus are rich mathematical objects possessig iterestig geometric, algebraic, topological ad
More informationWeek 56: The Binomial Coefficients
Wee 56: The Biomial Coefficiets March 6, 2018 1 Pascal Formula Theorem 11 (Pascal s Formula For itegers ad such that 1, ( ( ( 1 1 + 1 The umbers ( 2 ( 1 2 ( 2 are triagle umbers, that is, The petago umbers
More informationANSWERS SOLUTIONS iiii i. and 1. Thus, we have. i i i. i, A.
013 ΜΑΘ Natioal Covetio ANSWERS (1) C A A A B (6) B D D A B (11) C D D A A (16) D B A A C (1) D B C B C (6) D C B C C 1. We have SOLUTIONS 1 3 11 61 iiii 131161 i 013 013, C.. The powers of i cycle betwee
More informationRADICAL EXPRESSION. If a and x are real numbers and n is a positive integer, then x is an. n th root theorems: Example 1 Simplify
Example 1 Simplify 1.2A Radical Operatios a) 4 2 b) 16 1 2 c) 16 d) 2 e) 8 1 f) 8 What is the relatioship betwee a, b, c? What is the relatioship betwee d, e, f? If x = a, the x = = th root theorems: RADICAL
More informationSequences, Mathematical Induction, and Recursion. CSE 2353 Discrete Computational Structures Spring 2018
CSE 353 Discrete Computatioal Structures Sprig 08 Sequeces, Mathematical Iductio, ad Recursio (Chapter 5, Epp) Note: some course slides adopted from publisherprovided material Overview May mathematical
More informationSimple Polygons of Maximum Perimeter Contained in a Unit Disk
Discrete Comput Geom (009) 1: 08 15 DOI 10.1007/s00500890937 Simple Polygos of Maximum Perimeter Cotaied i a Uit Disk Charles Audet Pierre Hase Frédéric Messie Received: 18 September 007 / Revised:
More information1. By using truth tables prove that, for all statements P and Q, the statement
Author: Satiago Salazar Problems I: Mathematical Statemets ad Proofs. By usig truth tables prove that, for all statemets P ad Q, the statemet P Q ad its cotrapositive ot Q (ot P) are equivalet. I example.2.3
More informationPenrose Tilings and Periodicity
Perose Tiligs ad Periodicity Christopher McMurdie Mathematics Departmet, Uiversity of Arizoa 67 N. Sata Rita Ave., Box 0089 Tucso, AZ. 857 Email: mcmurdie@email.arizoa.edu Project: This is the first part
More informationSequences A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence
Sequeces A sequece of umbers is a fuctio whose domai is the positive itegers. We ca see that the sequece 1, 1, 2, 2, 3, 3,... is a fuctio from the positive itegers whe we write the first sequece elemet
More informationProblem Cosider the curve give parametrically as x = si t ad y = + cos t for» t» ß: (a) Describe the path this traverses: Where does it start (whe t =
Mathematics Summer Wilso Fial Exam August 8, ANSWERS Problem 1 (a) Fid the solutio to y +x y = e x x that satisfies y() = 5 : This is already i the form we used for a first order liear differetial equatio,
More informationChapter 1. Complex Numbers. Dr. Pulak Sahoo
Chapter 1 Complex Numbers BY Dr. Pulak Sahoo Assistat Professor Departmet of Mathematics Uiversity Of Kalyai West Begal, Idia Email : sahoopulak1@gmail.com 1 Module2: Stereographic Projectio 1 Euler
More informationA Simplified Binet Formula for kgeneralized Fibonacci Numbers
A Simplified Biet Formula for kgeeralized Fiboacci Numbers Gregory P. B. Dresde Departmet of Mathematics Washigto ad Lee Uiversity Lexigto, VA 440 dresdeg@wlu.edu Zhaohui Du Shaghai, Chia zhao.hui.du@gmail.com
More informationPolynomial Functions and Their Graphs
Polyomial Fuctios ad Their Graphs I this sectio we begi the study of fuctios defied by polyomial expressios. Polyomial ad ratioal fuctios are the most commo fuctios used to model data, ad are used extesively
More informationSubstitute these values into the first equation to get ( z + 6) + ( z + 3) + z = 27. Then solve to get
Problem ) The sum of three umbers is 7. The largest mius the smallest is 6. The secod largest mius the smallest is. What are the three umbers? [Problem submitted by Vi Lee, LCC Professor of Mathematics.
More informationProof of Fermat s Last Theorem by Algebra Identities and Linear Algebra
Proof of Fermat s Last Theorem by Algebra Idetities ad Liear Algebra Javad Babaee Ragai Youg Researchers ad Elite Club, Qaemshahr Brach, Islamic Azad Uiversity, Qaemshahr, Ira Departmet of Civil Egieerig,
More informationBINOMIAL PREDICTORS. + 2 j 1. Then n + 1 = The row of the binomial coefficients { ( n
BINOMIAL PREDICTORS VLADIMIR SHEVELEV arxiv:0907.3302v2 [math.nt] 22 Jul 2009 Abstract. For oegative itegers, k, cosider the set A,k = { [0, 1,..., ] : 2 k ( ). Let the biary epasio of + 1 be: + 1 = 2
More informationComplex Analysis Spring 2001 Homework I Solution
Complex Aalysis Sprig 2001 Homework I Solutio 1. Coway, Chapter 1, sectio 3, problem 3. Describe the set of poits satisfyig the equatio z a z + a = 2c, where c > 0 ad a R. To begi, we see from the triagle
More informationINEQUALITIES BJORN POONEN
INEQUALITIES BJORN POONEN 1 The AMGM iequality The most basic arithmetic meageometric mea (AMGM) iequality states simply that if x ad y are oegative real umbers, the (x + y)/2 xy, with equality if ad
More informationR is a scalar defined as follows:
Math 8. Notes o Dot Product, Cross Product, Plaes, Area, ad Volumes This lecture focuses primarily o the dot product ad its may applicatios, especially i the measuremet of agles ad scalar projectio ad
More informationSection 1.1. Calculus: Areas And Tangents. Difference Equations to Differential Equations
Differece Equatios to Differetial Equatios Sectio. Calculus: Areas Ad Tagets The study of calculus begis with questios about chage. What happes to the velocity of a swigig pedulum as its positio chages?
More informationAnalytic Continuation
Aalytic Cotiuatio The stadard example of this is give by Example Let h (z) = 1 + z + z 2 + z 3 +... kow to coverge oly for z < 1. I fact h (z) = 1/ (1 z) for such z. Yet H (z) = 1/ (1 z) is defied for
More informationSolutions to Homework 1
Solutios to Homework MATH 36. Describe geometrically the sets of poits z i the complex plae defied by the followig relatios /z = z () Re(az + b) >, where a, b (2) Im(z) = c, with c (3) () = = z z = z 2.
More information(a) (b) All real numbers. (c) All real numbers. (d) None. to show the. (a) 3. (b) [ 7, 1) (c) ( 7, 1) (d) At x = 7. (a) (b)
Chapter 0 Review 597. E; a ( + )( + ) + + S S + S + + + + + + S lim + l. D; a diverges by the Itegral l k Test sice d lim [(l ) ], so k l ( ) does ot coverge absolutely. But it coverges by the Alteratig
More informationROSE WONG. f(1) f(n) where L the average value of f(n). In this paper, we will examine averages of several different arithmetic functions.
AVERAGE VALUES OF ARITHMETIC FUNCTIONS ROSE WONG Abstract. I this paper, we will preset problems ivolvig average values of arithmetic fuctios. The arithmetic fuctios we discuss are: (1)the umber of represetatios
More informationWe are mainly going to be concerned with power series in x, such as. (x)} converges  that is, lims N n
Review of Power Series, Power Series Solutios A power series i x  a is a ifiite series of the form c (x a) =c +c (x a)+(x a) +... We also call this a power series cetered at a. Ex. (x+) is cetered at
More information18th Bay Area Mathematical Olympiad. Problems and Solutions. February 23, 2016
18th Bay Area Mathematical Olympiad February 3, 016 Problems ad Solutios BAMO8 ad BAMO1 are each 5questio essayproof exams, for middle ad highschool studets, respectively. The problems i each exam
More informationn m CHAPTER 3 RATIONAL EXPONENTS AND RADICAL FUNCTIONS 31 Evaluate n th Roots and Use Rational Exponents Real nth Roots of a n th Root of a
CHAPTER RATIONAL EXPONENTS AND RADICAL FUNCTIONS Big IDEAS: 1) Usig ratioal expoets ) Performig fuctio operatios ad fidig iverse fuctios ) Graphig radical fuctios ad solvig radical equatios Sectio: Essetial
More informationABOUT CHAOS AND SENSITIVITY IN TOPOLOGICAL DYNAMICS
ABOUT CHAOS AND SENSITIVITY IN TOPOLOGICAL DYNAMICS EDUARD KONTOROVICH Abstract. I this work we uify ad geeralize some results about chaos ad sesitivity. Date: March 1, 005. 1 1. Symbolic Dyamics Defiitio
More informationA NOTE ON INVARIANT SETS OF ITERATED FUNCTION SYSTEMS
Acta Math. Hugar., 2007 DOI: 10.1007/s1047400770136 A NOTE ON INVARIANT SETS OF ITERATED FUNCTION SYSTEMS L. L. STACHÓ ad L. I. SZABÓ Bolyai Istitute, Uiversity of Szeged, Aradi vértaúk tere 1, H6720
More informationIn number theory we will generally be working with integers, though occasionally fractions and irrationals will come into play.
Number Theory Math 5840 otes. Sectio 1: Axioms. I umber theory we will geerally be workig with itegers, though occasioally fractios ad irratioals will come ito play. Notatio: Z deotes the set of all itegers
More informationSEQUENCES AND SERIES
9 SEQUENCES AND SERIES INTRODUCTION Sequeces have may importat applicatios i several spheres of huma activities Whe a collectio of objects is arraged i a defiite order such that it has a idetified first
More informationMathematical Induction
Mathematical Iductio Itroductio Mathematical iductio, or just iductio, is a proof techique. Suppose that for every atural umber, P() is a statemet. We wish to show that all statemets P() are true. I a
More information1 Generating functions for balls in boxes
Math 566 Fall 05 Some otes o geeratig fuctios Give a sequece a 0, a, a,..., a,..., a geeratig fuctio some way of represetig the sequece as a fuctio. There are may ways to do this, with the most commo ways
More informationMATH 205 HOMEWORK #2 OFFICIAL SOLUTION. (f + g)(x) = f(x) + g(x) = f( x) g( x) = (f + g)( x)
MATH 205 HOMEWORK #2 OFFICIAL SOLUTION Problem 2: Do problems 79 o page 40 of Hoffma & Kuze. (7) We will prove this by cotradictio. Suppose that W 1 is ot cotaied i W 2 ad W 2 is ot cotaied i W 1. The
More information3.2 Properties of Division 3.3 Zeros of Polynomials 3.4 Complex and Rational Zeros of Polynomials
Math 60 www.timetodare.com 3. Properties of Divisio 3.3 Zeros of Polyomials 3.4 Complex ad Ratioal Zeros of Polyomials I these sectios we will study polyomials algebraically. Most of our work will be cocered
More informationMA131  Analysis 1. Workbook 2 Sequences I
MA3  Aalysis Workbook 2 Sequeces I Autum 203 Cotets 2 Sequeces I 2. Itroductio.............................. 2.2 Icreasig ad Decreasig Sequeces................ 2 2.3 Bouded Sequeces..........................
More informationCSE 1400 Applied Discrete Mathematics Number Theory and Proofs
CSE 1400 Applied Discrete Mathematics Number Theory ad Proofs Departmet of Computer Scieces College of Egieerig Florida Tech Sprig 01 Problems for Number Theory Backgroud Number theory is the brach of
More informationA sequence of numbers is a function whose domain is the positive integers. We can see that the sequence
Sequeces A sequece of umbers is a fuctio whose domai is the positive itegers. We ca see that the sequece,, 2, 2, 3, 3,... is a fuctio from the positive itegers whe we write the first sequece elemet as
More informationOn Random Line Segments in the Unit Square
O Radom Lie Segmets i the Uit Square Thomas A. Courtade Departmet of Electrical Egieerig Uiversity of Califoria Los Ageles, Califoria 90095 Email: tacourta@ee.ucla.edu I. INTRODUCTION Let Q = [0, 1] [0,
More informationZeros of Polynomials
Math 160 www.timetodare.com 4.5 4.6 Zeros of Polyomials I these sectios we will study polyomials algebraically. Most of our work will be cocered with fidig the solutios of polyomial equatios of ay degree
More informationInfinite Sequences and Series
Chapter 6 Ifiite Sequeces ad Series 6.1 Ifiite Sequeces 6.1.1 Elemetary Cocepts Simply speakig, a sequece is a ordered list of umbers writte: {a 1, a 2, a 3,...a, a +1,...} where the elemets a i represet
More informationComparison Study of Series Approximation. and Convergence between Chebyshev. and Legendre Series
Applied Mathematical Scieces, Vol. 7, 03, o. 6, 3337 HIKARI Ltd, www.mhikari.com http://d.doi.org/0.988/ams.03.3430 Compariso Study of Series Approimatio ad Covergece betwee Chebyshev ad Legedre Series
More informationPatterns in Complex Numbers An analytical paper on the roots of a complex numbers and its geometry
IB MATHS HL POTFOLIO TYPE Patters i Complex Numbers A aalytical paper o the roots of a complex umbers ad its geometry i Syed Tousif Ahmed Cadidate Sessio Number: 0066009 School Code: 0066 Sessio: May
More informationNICK DUFRESNE. 1 1 p(x). To determine some formulas for the generating function of the Schröder numbers, r(x) = a(x) =
AN INTRODUCTION TO SCHRÖDER AND UNKNOWN NUMBERS NICK DUFRESNE Abstract. I this article we will itroduce two types of lattice paths, Schröder paths ad Ukow paths. We will examie differet properties of each,
More informationSolutions for May. 3 x + 7 = 4 x x +
Solutios for May 493. Prove that there is a atural umber with the followig characteristics: a) it is a multiple of 007; b) the first four digits i its decimal represetatio are 009; c) the last four digits
More informationSequences and Series of Functions
Chapter 6 Sequeces ad Series of Fuctios 6.1. Covergece of a Sequece of Fuctios Poitwise Covergece. Defiitio 6.1. Let, for each N, fuctio f : A R be defied. If, for each x A, the sequece (f (x)) coverges
More informationIf a subset E of R contains no open interval, is it of zero measure? For instance, is the set of irrationals in [0, 1] is of measure zero?
2 Lebesgue Measure I Chapter 1 we defied the cocept of a set of measure zero, ad we have observed that every coutable set is of measure zero. Here are some atural questios: If a subset E of R cotais a
More informationThe OnLine Heilbronn s Triangle Problem in d Dimensions
Discrete Comput Geom 38:5 60 2007 DI: 0.007/s00454007323x Discrete & Computatioal Geometry 2007 Spriger Sciece+Busiess Media, Ic. The Lie Heilbro s Triagle Problem i d Dimesios Gill Barequet ad Alia
More information2 Geometric interpretation of complex numbers
2 Geometric iterpretatio of complex umbers 2.1 Defiitio I will start fially with a precise defiitio, assumig that such mathematical object as vector space R 2 is well familiar to the studets. Recall that
More informationSEQUENCE AND SERIES NCERT
9. Overview By a sequece, we mea a arragemet of umbers i a defiite order accordig to some rule. We deote the terms of a sequece by a, a,..., etc., the subscript deotes the positio of the term. I view of
More informationMath 220A Fall 2007 Homework #2. Will Garner A
Math 0A Fall 007 Homewor # Will Garer Pg 3 #: Show that {cis : a oegative iteger} is dese i T = {z œ : z = }. For which values of q is {cis(q): a oegative iteger} dese i T? To show that {cis : a oegative
More information3. Z Transform. Recall that the Fourier transform (FT) of a DT signal xn [ ] is ( ) [ ] = In order for the FT to exist in the finite magnitude sense,
3. Z Trasform Referece: Etire Chapter 3 of text. Recall that the Fourier trasform (FT) of a DT sigal x [ ] is ω ( ) [ ] X e = j jω k = xe I order for the FT to exist i the fiite magitude sese, S = x [
More informationAPPENDIX F Complex Numbers
APPENDIX F Complex Numbers Operatios with Complex Numbers Complex Solutios of Quadratic Equatios Polar Form of a Complex Number Powers ad Roots of Complex Numbers Operatios with Complex Numbers Some equatios
More information[ 11 ] z of degree 2 as both degree 2 each. The degree of a polynomial in n variables is the maximum of the degrees of its terms.
[ 11 ] 1 1.1 Polyomial Fuctios 1 Algebra Ay fuctio f ( x) ax a1x... a1x a0 is a polyomial fuctio if ai ( i 0,1,,,..., ) is a costat which belogs to the set of real umbers ad the idices,, 1,...,1 are atural
More informationObjective Mathematics
. If sum of '' terms of a sequece is give by S Tr ( )( ), the 4 5 67 r (d) 4 9 r is equal to : T. Let a, b, c be distict ozero real umbers such that a, b, c are i harmoic progressio ad a, b, c are i arithmetic
More informationExercises 1 Sets and functions
Exercises 1 Sets ad fuctios HU Wei September 6, 018 1 Basics Set theory ca be made much more rigorous ad built upo a set of Axioms. But we will cover oly some heuristic ideas. For those iterested studets,
More informationCHAPTER 10 INFINITE SEQUENCES AND SERIES
CHAPTER 10 INFINITE SEQUENCES AND SERIES 10.1 Sequeces 10.2 Ifiite Series 10.3 The Itegral Tests 10.4 Compariso Tests 10.5 The Ratio ad Root Tests 10.6 Alteratig Series: Absolute ad Coditioal Covergece
More informationSequences I. Chapter Introduction
Chapter 2 Sequeces I 2. Itroductio A sequece is a list of umbers i a defiite order so that we kow which umber is i the first place, which umber is i the secod place ad, for ay atural umber, we kow which
More informationThe multiplicative structure of finite field and a construction of LRC
IERG6120 Codig for Distributed Storage Systems Lecture 806/10/2016 The multiplicative structure of fiite field ad a costructio of LRC Lecturer: Keeth Shum Scribe: Zhouyi Hu Notatios: We use the otatio
More information6.3 Testing Series With Positive Terms
6.3. TESTING SERIES WITH POSITIVE TERMS 307 6.3 Testig Series With Positive Terms 6.3. Review of what is kow up to ow I theory, testig a series a i for covergece amouts to fidig the i= sequece of partial
More informationMath 475, Problem Set #12: Answers
Math 475, Problem Set #12: Aswers A. Chapter 8, problem 12, parts (b) ad (d). (b) S # (, 2) = 2 2, sice, from amog the 2 ways of puttig elemets ito 2 distiguishable boxes, exactly 2 of them result i oe
More informationWe will conclude the chapter with the study a few methods and techniques which are useful
Chapter : Coordiate geometry: I this chapter we will lear about the mai priciples of graphig i a dimesioal (D) Cartesia system of coordiates. We will focus o drawig lies ad the characteristics of the graphs
More informationUNIVERSITY OF NORTHERN COLORADO MATHEMATICS CONTEST. First Round For all Colorado Students Grades 712 November 3, 2007
UNIVERSITY OF NORTHERN COLORADO MATHEMATICS CONTEST First Roud For all Colorado Studets Grades 7 November, 7 The positive itegers are,,, 4, 5, 6, 7, 8, 9,,,,. The Pythagorea Theorem says that a + b =
More informationAlternating Series. 1 n 0 2 n n THEOREM 9.14 Alternating Series Test Let a n > 0. The alternating series. 1 n a n.
0_0905.qxd //0 :7 PM Page SECTION 9.5 Alteratig Series Sectio 9.5 Alteratig Series Use the Alteratig Series Test to determie whether a ifiite series coverges. Use the Alteratig Series Remaider to approximate
More informationLONG SNAKES IN POWERS OF THE COMPLETE GRAPH WITH AN ODD NUMBER OF VERTICES
J Lodo Math Soc (2 50, (1994, 465 476 LONG SNAKES IN POWERS OF THE COMPLETE GRAPH WITH AN ODD NUMBER OF VERTICES Jerzy Wojciechowski Abstract I [5] Abbott ad Katchalski ask if there exists a costat c >
More informationWHEN IS THE (CO)SINE OF A RATIONAL ANGLE EQUAL TO A RATIONAL NUMBER? 1. My Motivation Some Sort of an Introduction
WHEN IS THE (CO)SINE OF A RATIONAL ANGLE EQUAL TO A RATIONAL NUMBER? JÖRG JAHNEL 1. My Motivatio Some Sort of a Itroductio Last term I taught Topological Groups at the Göttige Georg August Uiversity. This
More informationArea As A Limit & Sigma Notation
Area As A Limit & Sigma Notatio SUGGESTED REFERENCE MATERIAL: As you work through the problems listed below, you should referece Chapter 5.4 of the recommeded textbook (or the equivalet chapter i your
More informationCHAPTER I: Vector Spaces
CHAPTER I: Vector Spaces Sectio 1: Itroductio ad Examples This first chapter is largely a review of topics you probably saw i your liear algebra course. So why cover it? (1) Not everyoe remembers everythig
More informationSolving equations (incl. radical equations) involving these skills, but ultimately solvable by factoring/quadratic formula (no complex roots)
Evet A: Fuctios ad Algebraic Maipulatio Factorig Square of a sum: ( a + b) = a + ab + b Square of a differece: ( a b) = a ab + b Differece of squares: a b = ( a b )(a + b ) Differece of cubes: a 3 b 3
More informationNBHM QUESTION 2007 Section 1 : Algebra Q1. Let G be a group of order n. Which of the following conditions imply that G is abelian?
NBHM QUESTION 7 NBHM QUESTION 7 NBHM QUESTION 7 Sectio : Algebra Q Let G be a group of order Which of the followig coditios imply that G is abelia? 5 36 Q Which of the followig subgroups are ecesarily
More informationFundamental Concepts: Surfaces and Curves
UNDAMENTAL CONCEPTS: SURACES AND CURVES CHAPTER udametal Cocepts: Surfaces ad Curves. INTRODUCTION This chapter describes two geometrical objects, vi., surfaces ad curves because the pla a ver importat
More informationChapter 6 Infinite Series
Chapter 6 Ifiite Series I the previous chapter we cosidered itegrals which were improper i the sese that the iterval of itegratio was ubouded. I this chapter we are goig to discuss a topic which is somewhat
More informationContinuous Functions
Cotiuous Fuctios Q What does it mea for a fuctio to be cotiuous at a poit? Aswer I mathematics, we have a defiitio that cosists of three cocepts that are liked i a special way Cosider the followig defiitio
More informationMath 142, Final Exam. 5/2/11.
Math 4, Fial Exam 5// No otes, calculator, or text There are poits total Partial credit may be give Write your full ame i the upper right corer of page Number the pages i the upper right corer Do problem
More informationNAME: ALGEBRA 350 BLOCK 7. Simplifying Radicals Packet PART 1: ROOTS
NAME: ALGEBRA 50 BLOCK 7 DATE: Simplifyig Radicals Packet PART 1: ROOTS READ: A square root of a umber b is a solutio of the equatio x = b. Every positive umber b has two square roots, deoted b ad b or
More informationON THE LEHMER CONSTANT OF FINITE CYCLIC GROUPS
ON THE LEHMER CONSTANT OF FINITE CYCLIC GROUPS NORBERT KAIBLINGER Abstract. Results of Lid o Lehmer s problem iclude the value of the Lehmer costat of the fiite cyclic group Z/Z, for 5 ad all odd. By complemetary
More informationPresentation of complex number in Cartesian and polar coordinate system
a + bi, aεr, bεr i = z = a + bi a = Re(z), b = Im(z) give z = a + bi & w = c + di, a + bi = c + di a = c & b = d The complex cojugate of z = a + bi is z = a bi The sum of complex cojugates is real: z +
More informationSome sufficient conditions of a given. series with rational terms converging to an irrational number or a transcdental number
Some sufficiet coditios of a give arxiv:0807.376v2 [math.nt] 8 Jul 2008 series with ratioal terms covergig to a irratioal umber or a trascdetal umber Yu Gao,Jiig Gao Shaghai Putuo college, Shaghai Jiaotog
More informationOrthogonal transformations
Orthogoal trasformatios October 12, 2014 1 Defiig property The squared legth of a vector is give by takig the dot product of a vector with itself, v 2 v v g ij v i v j A orthogoal trasformatio is a liear
More informationREGULARIZATION OF CERTAIN DIVERGENT SERIES OF POLYNOMIALS
REGULARIZATION OF CERTAIN DIVERGENT SERIES OF POLYNOMIALS LIVIU I. NICOLAESCU ABSTRACT. We ivestigate the geeralized covergece ad sums of series of the form P at P (x, where P R[x], a R,, ad T : R[x] R[x]
More information